n identical condensers are joined in parallel and are charged to potential so that energy stored in cach condenser is . If they are seperated and joined in series, then the total energy and total potential difference of the combination will be

Options

A. ` nE andV/E`
B. `n^2E and nV`
C. `E/(n^2) and V/(n^2)`
D. nE and nV

Answer

To solve the problem step by step, we will analyze the situation of capacitors connected in parallel and then in series.

### Step 1: Understand the Energy in Parallel Configuration
When n identical capacitors are connected in parallel, the energy stored in each capacitor is given as \( E \). The total energy stored in the parallel configuration can be expressed as:
\[ E_{total} = nE \]
where \( E \) is the energy stored in each capacitor.

**Hint:** Remember that in a parallel configuration, the voltage across each capacitor is the same, and the total energy is the sum of the energy stored in each capacitor.

### Step 2: Determine the Voltage Across Each Capacitor
Let the voltage across each capacitor in the parallel configuration be \( V \). The energy stored in a single capacitor is given by:
\[ E = \frac{1}{2} C V^2 \]
where \( C \) is the capacitance of each capacitor.

**Hint:** Use the formula for energy stored in a capacitor to relate energy to voltage and capacitance.

### Step 3: Transition to Series Configuration
When the capacitors are separated and connected in series, the total voltage across the series combination becomes:
\[ V_{total} = nV \]
because the voltages across each capacitor add up in series.

**Hint:** In a series configuration, the total voltage is the sum of the voltages across each individual capacitor.

### Step 4: Calculate the Equivalent Capacitance in Series
The equivalent capacitance \( C_{eq} \) for n identical capacitors in series is given by:
\[ \frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{C} + \ldots + \frac{1}{C} = \frac{n}{C} \]
Thus, the equivalent capacitance is:
\[ C_{eq} = \frac{C}{n} \]

**Hint:** Remember the formula for equivalent capacitance in series and how it changes with the number of capacitors.

### Step 5: Calculate the Energy in Series Configuration
The energy stored in the series configuration can be calculated using the equivalent capacitance and the total voltage:
\[ E_{total, series} = \frac{1}{2} C_{eq} V_{total}^2 \]
Substituting the values we found:
\[ E_{total, series} = \frac{1}{2} \left(\frac{C}{n}\right) (nV)^2 \]
\[ E_{total, series} = \frac{1}{2} \left(\frac{C}{n}\right) n^2 V^2 \]
\[ E_{total, series} = \frac{n}{2} C V^2 \]

### Step 6: Relate the Energy in Series to the Original Energy
From the expression for energy stored in a single capacitor, we have:
\[ E = \frac{1}{2} C V^2 \]
Thus, we can express the energy in the series configuration as:
\[ E_{total, series} = nE \]

### Final Result
The total energy stored in the series configuration is:
\[ E_{total, series} = nE \]
And the total potential difference across the series combination is:
\[ V_{total} = nV \]

### Conclusion
The total energy and total potential difference of the combination when the capacitors are connected in series is:
- Total Energy: \( nE \)
- Total Potential Difference: \( nV \)